3.33.98 \(\int \frac {-5+12 x+12 x^2-4 x^3+e^{4+x} (-20 x-4 x^2+2 x^3)}{e^4} \, dx\)

Optimal. Leaf size=24 \[ \frac {(5-x) x \left (-1+x+x \left (-2 e^{4+x}+x\right )\right )}{e^4} \]

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Rubi [A]  time = 0.11, antiderivative size = 47, normalized size of antiderivative = 1.96, number of steps used = 14, number of rules used = 5, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {12, 1594, 2196, 2176, 2194} \begin {gather*} -\frac {x^4}{e^4}+2 e^x x^3+\frac {4 x^3}{e^4}-10 e^x x^2+\frac {6 x^2}{e^4}-\frac {5 x}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + 12*x + 12*x^2 - 4*x^3 + E^(4 + x)*(-20*x - 4*x^2 + 2*x^3))/E^4,x]

[Out]

(-5*x)/E^4 + (6*x^2)/E^4 - 10*E^x*x^2 + (4*x^3)/E^4 + 2*E^x*x^3 - x^4/E^4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (-5+12 x+12 x^2-4 x^3+e^{4+x} \left (-20 x-4 x^2+2 x^3\right )\right ) \, dx}{e^4}\\ &=-\frac {5 x}{e^4}+\frac {6 x^2}{e^4}+\frac {4 x^3}{e^4}-\frac {x^4}{e^4}+\frac {\int e^{4+x} \left (-20 x-4 x^2+2 x^3\right ) \, dx}{e^4}\\ &=-\frac {5 x}{e^4}+\frac {6 x^2}{e^4}+\frac {4 x^3}{e^4}-\frac {x^4}{e^4}+\frac {\int e^{4+x} x \left (-20-4 x+2 x^2\right ) \, dx}{e^4}\\ &=-\frac {5 x}{e^4}+\frac {6 x^2}{e^4}+\frac {4 x^3}{e^4}-\frac {x^4}{e^4}+\frac {\int \left (-20 e^{4+x} x-4 e^{4+x} x^2+2 e^{4+x} x^3\right ) \, dx}{e^4}\\ &=-\frac {5 x}{e^4}+\frac {6 x^2}{e^4}+\frac {4 x^3}{e^4}-\frac {x^4}{e^4}+\frac {2 \int e^{4+x} x^3 \, dx}{e^4}-\frac {4 \int e^{4+x} x^2 \, dx}{e^4}-\frac {20 \int e^{4+x} x \, dx}{e^4}\\ &=-\frac {5 x}{e^4}-20 e^x x+\frac {6 x^2}{e^4}-4 e^x x^2+\frac {4 x^3}{e^4}+2 e^x x^3-\frac {x^4}{e^4}-\frac {6 \int e^{4+x} x^2 \, dx}{e^4}+\frac {8 \int e^{4+x} x \, dx}{e^4}+\frac {20 \int e^{4+x} \, dx}{e^4}\\ &=20 e^x-\frac {5 x}{e^4}-12 e^x x+\frac {6 x^2}{e^4}-10 e^x x^2+\frac {4 x^3}{e^4}+2 e^x x^3-\frac {x^4}{e^4}-\frac {8 \int e^{4+x} \, dx}{e^4}+\frac {12 \int e^{4+x} x \, dx}{e^4}\\ &=12 e^x-\frac {5 x}{e^4}+\frac {6 x^2}{e^4}-10 e^x x^2+\frac {4 x^3}{e^4}+2 e^x x^3-\frac {x^4}{e^4}-\frac {12 \int e^{4+x} \, dx}{e^4}\\ &=-\frac {5 x}{e^4}+\frac {6 x^2}{e^4}-10 e^x x^2+\frac {4 x^3}{e^4}+2 e^x x^3-\frac {x^4}{e^4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 23, normalized size = 0.96 \begin {gather*} -\frac {(-5+x) x \left (-1+x-2 e^{4+x} x+x^2\right )}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 12*x + 12*x^2 - 4*x^3 + E^(4 + x)*(-20*x - 4*x^2 + 2*x^3))/E^4,x]

[Out]

-(((-5 + x)*x*(-1 + x - 2*E^(4 + x)*x + x^2))/E^4)

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fricas [A]  time = 0.63, size = 36, normalized size = 1.50 \begin {gather*} -{\left (x^{4} - 4 \, x^{3} - 6 \, x^{2} - 2 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{\left (x + 4\right )} + 5 \, x\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-4*x^2-20*x)*exp(4+x)-4*x^3+12*x^2+12*x-5)/exp(4),x, algorithm="fricas")

[Out]

-(x^4 - 4*x^3 - 6*x^2 - 2*(x^3 - 5*x^2)*e^(x + 4) + 5*x)*e^(-4)

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giac [A]  time = 0.31, size = 36, normalized size = 1.50 \begin {gather*} -{\left (x^{4} - 4 \, x^{3} - 6 \, x^{2} - 2 \, {\left (x^{3} - 5 \, x^{2}\right )} e^{\left (x + 4\right )} + 5 \, x\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-4*x^2-20*x)*exp(4+x)-4*x^3+12*x^2+12*x-5)/exp(4),x, algorithm="giac")

[Out]

-(x^4 - 4*x^3 - 6*x^2 - 2*(x^3 - 5*x^2)*e^(x + 4) + 5*x)*e^(-4)

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maple [A]  time = 0.06, size = 42, normalized size = 1.75

method	result	size
risch	\(-{\mathrm e}^{-4} x^{4}+4 \,{\mathrm e}^{-4} x^{3}+6 x^{2} {\mathrm e}^{-4}-5 x \,{\mathrm e}^{-4}+\left (2 x^{3}-10 x^{2}\right ) {\mathrm e}^{x}\)	\(42\)
default	\({\mathrm e}^{-4} \left (-5 x +2 \,{\mathrm e}^{4+x} \left (4+x \right )^{3}-34 \,{\mathrm e}^{4+x} \left (4+x \right )^{2}+176 \,{\mathrm e}^{4+x} \left (4+x \right )-288 \,{\mathrm e}^{4+x}+6 x^{2}+4 x^{3}-x^{4}\right )\)	\(62\)
norman	\(-5 x \,{\mathrm e}^{-4}+6 x^{2} {\mathrm e}^{-4}+4 \,{\mathrm e}^{-4} x^{3}-{\mathrm e}^{-4} x^{4}-10 x^{2} {\mathrm e}^{-4} {\mathrm e}^{4+x}+2 \,{\mathrm e}^{-4} x^{3} {\mathrm e}^{4+x}\)	\(62\)
derivativedivides	\({\mathrm e}^{-4} \left (-212-53 x +2 \,{\mathrm e}^{4+x} \left (4+x \right )^{3}-34 \,{\mathrm e}^{4+x} \left (4+x \right )^{2}+176 \,{\mathrm e}^{4+x} \left (4+x \right )-288 \,{\mathrm e}^{4+x}+6 \left (4+x \right )^{2}+4 x^{3}-x^{4}\right )\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3-4*x^2-20*x)*exp(4+x)-4*x^3+12*x^2+12*x-5)/exp(4),x,method=_RETURNVERBOSE)

[Out]

-exp(-4)*x^4+4*exp(-4)*x^3+6*x^2*exp(-4)-5*x*exp(-4)+(2*x^3-10*x^2)*exp(x)

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maxima [A]  time = 0.39, size = 39, normalized size = 1.62 \begin {gather*} -{\left (x^{4} - 4 \, x^{3} - 6 \, x^{2} - 2 \, {\left (x^{3} e^{4} - 5 \, x^{2} e^{4}\right )} e^{x} + 5 \, x\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-4*x^2-20*x)*exp(4+x)-4*x^3+12*x^2+12*x-5)/exp(4),x, algorithm="maxima")

[Out]

-(x^4 - 4*x^3 - 6*x^2 - 2*(x^3*e^4 - 5*x^2*e^4)*e^x + 5*x)*e^(-4)

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mupad [B]  time = 2.05, size = 21, normalized size = 0.88 \begin {gather*} -x\,{\mathrm {e}}^{-4}\,\left (x-5\right )\,\left (x-2\,x\,{\mathrm {e}}^{x+4}+x^2-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-4)*(exp(x + 4)*(20*x + 4*x^2 - 2*x^3) - 12*x - 12*x^2 + 4*x^3 + 5),x)

[Out]

-x*exp(-4)*(x - 5)*(x - 2*x*exp(x + 4) + x^2 - 1)

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sympy [B]  time = 0.15, size = 48, normalized size = 2.00 \begin {gather*} - \frac {x^{4}}{e^{4}} + \frac {4 x^{3}}{e^{4}} + \frac {6 x^{2}}{e^{4}} - \frac {5 x}{e^{4}} + \frac {\left (2 x^{3} - 10 x^{2}\right ) e^{x + 4}}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3-4*x**2-20*x)*exp(4+x)-4*x**3+12*x**2+12*x-5)/exp(4),x)

[Out]

-x**4*exp(-4) + 4*x**3*exp(-4) + 6*x**2*exp(-4) - 5*x*exp(-4) + (2*x**3 - 10*x**2)*exp(-4)*exp(x + 4)

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